Let $f$ be a meromorphic function on a Riemann surface $X$ , with associated holomorphic map $F:X\rightarrow \mathbb{C_{\infty}}$.
Prove that :
a) If $p$ is zero of $f$ then $mult_p(F)=ord_p(f)$.
b)If $p$ is a pole of $f$ then $mult_p(F)=-ord_p(f)$.
This a lemma from Rick Miranda ' Algebraic curves and Riemann surfaces.
I want to verify my idea:
a)Define $F(x)= f(x) $ if $x $ is not a pole
and $F(x)=\infty$ if $x$ is a pole.
We know that order of a meromorphic function at a point on a Riemann surface is invariant of charts. Now take two charts around $p$ , and $F(p)=f(p)=0$ say $(U,\phi)$ and $(V,I_{\mathbb{C}})$ respectively such that $I_{\mathbb{C}}\circ{F}\circ \phi^{-1}=F\circ \phi^{-1}$ is holomorphic and power series representation of $F\circ \phi^{-1}(z)-F\circ \phi^{-1}(\phi(p))=F\circ \phi^{-1}(z)-0=F\circ \phi^{-1}(z)$ around $\phi(p)$ has lowest degree say $m$ and $mult_p(F)=m$.
Now take take the same chart $(U,\phi)$ for $f$ at $p$ and $f\circ\phi^{-1}$ is has Laurent series expansion at $\phi(p)$ and $ord_p(f)>0$ , observe that $f\circ\phi^{-1}(z)= f(\phi^{-1}(z))=F(\phi^{-1}(z))\forall z\in\phi^{-1}(U) $ (I have doubt here).
therefore $Ord_p(f)=mult_p(F)$.
b) if $p$ is pole of $f$ then $F(p)=\infty$ and $p$ is a zero of $1/f$ now $ord_p(1/f)=-ord_p(f)$. And for $mult_p(F)$ we will take the chat around $p$ and $F(p)=\infty$ say $(U,\phi)$ and $(V,1/z)$ then power series of $\frac{1}{F\circ\phi^{-1}}$ is same as the Laurent series of $\frac{1}{f\circ\phi^{-1}}$
Therefore $mult_p(F)=ord_p(1/f)=-ord_p(f)$.
Hence the proof.
Best Answer
Part b) seems fine. Here is how I would argue with regards to part a).
Let $p\in X$ be a zero of $f$. Pick a chart $\phi\colon U \rightarrow V \subseteq \mathbb{C}$ of $X$ with $p \in U$. Since the singularities of a meromorphic function are isolated by definition (and $p$ is not a singularity), we can without loss of generality assume that $f$ is holomorphic on $U$. Thus the maps $f$ and $F$ agree on $U$. Hence, the maps $(\operatorname{id}_{\mathbb{C}}\circ F\circ \phi^{-1})$ and $(f\circ\phi^{-1})$ agree on $V$.
By the uniqueness of the Laurent expansion, the Taylor expansion and the Laurent expansion of the holomorphic function $(\operatorname{id}_{\mathbb{C}}\circ F\circ \phi^{-1})\restriction_V=(f\circ\phi^{-1})\restriction_V$ around $z_0\colon=\phi(p)$ are equal. Thus, if $k$ denotes the order of $f$ at $p$, then the Taylor expansion of $\operatorname{id}_{\mathbb{C}}\circ F\circ \phi^{-1}$ near $z_0$ is of the form $\sum_{n=k}^\infty c_n(z-z_0)^n$, where $c_n\in \mathbb{C}$. By Lemma 4.4 of Miranda's Algebraic Curves and Riemann Surfaces, the multiplicity of $F$ at $p$ is given by the exponent of the first strictly positive term of the Taylor expansion of $(\operatorname{id}_{\mathbb{C}}\circ F\circ \phi^{-1})\restriction_{V}$. With $k>0$ (as $f(p)=0$), the equality $\operatorname{mult}_p(F)=k$ follows.
One can easily generalize the statement of part a) as follows:
By part a), we have $\operatorname{ord}_p(f-f(p))=\operatorname{mult}_p(F-F(p))$. With $\operatorname{mult}_p(F-F(p))= \operatorname{mult}_p(F)$, the claim follows. To see the last equality, pick a coordinate neighborhood $(U,\phi)$ of $p$. Consider the chart $\psi:\mathbb{C}\rightarrow \mathbb{C};z\mapsto z-F(p)$ centered at $F(p)\in \mathbb{C}_\infty$. Then the Taylor expansions of $(F-F(p))\circ \phi^{-1}$ and $\psi\circ F\circ \phi^{-1}$ (both around $\phi(p)\in \mathbb{C}$) agree. Now, again use Lemma 4.4. of Miranda's Algebraic Curves and Riemann Surfaces.